Question: Let $A,$ $B,$ $C$ be the angles of a non-right triangle.  Compute
\[\begin{vmatrix} \tan A & 1  & 1 \\ 1 & \tan B & 1 \\ 1 & 1 & \tan C \end{vmatrix}.\]
Explanation: Expanding the determinant, we get
\begin{align*}
\begin{vmatrix} \tan 1 & 1  & 1 \\ 1 & \tan B & 1 \\ 1 & 1 & \tan C \end{vmatrix} &= \tan A \begin{vmatrix} \tan B & 1 \\ 1 & \tan C \end{vmatrix} - \begin{vmatrix} 1  & 1 \\ 1 & \tan C \end{vmatrix} + \begin{vmatrix} 1 & \tan B \\ 1 & 1 \end{vmatrix} \\
&= \tan A(\tan B \tan C - 1) - (\tan C - 1) + (1 - \tan B) \\
&= \tan A \tan B \tan C - \tan A - \tan B - \tan C + 2.
\end{align*}From the tangent addition formula,
\[\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}.\]But
\[\tan (A + B) = \tan (180^\circ - C) = -\tan C,\]so
\[-\tan C = \frac{\tan A + \tan B}{1 - \tan A \tan B}.\]Then $-\tan C + \tan A \tan B \tan C = \tan A + \tan B.$  Therefore,
\[\tan A \tan B \tan C - \tan A - \tan B - \tan C + 2 = \boxed{2}.\]